MCQ
A transition metal $M$ forms a volatile chloride which has a vapour density of $94.8$. If it contains $74.75\%$ of chlorine the formula of the metal chloride will be
- A$MCl_3$
- B$MCl_2$
- ✓$MCl_4$
- D$MCl_5$
Weight of metal $= 100\,g - 74.75\,g$
$= 25.25\,g$
Equivalent weight
$ = \frac{{weight\,of\,metal}}{{weight\,of\,chlorine\,}} \times 35.5$
$ = \frac{{25.25}}{{74.75}} \times 35.5 = 12$
Valency of metal
$ = \frac{{2 \times V.D.}}{{Equivalent\,wt.of\,metal + 35.5}}$
$ = \frac{{2 \times 94.8}}{{12 + 35.5}} = 4$
$\therefore $ Formula of compound $= MCl_4$
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$\left[\right.$ Given : $pK _{ b }\left( NH _3\right)=4.745$
$\log 2=0.301$
$\log 3=0.477$
$T =298\,K ]$